Integral exercise asked by Kunihiko Chikaya in many math groups Level 3

Exercise:

Compute, $\int_{0}^{\frac{\pi }{2}}{\frac{dx}{{{\left( \sin \left( a+x \right)+\cos x \right)}^{2}}}\,\,\,,\,\,\,\,\left| a \right|<\frac{\pi }{2}}$

Solution: we know that $\cos x=\sin \left( \frac{\pi }{2}-x \right)$

So $\sin \left( a+x \right)+\cos x=\sin \left( a+x \right)+\sin \left( \frac{\pi }{2}-x \right)$

Now using $\sin p+\sin q=2\sin \left( \frac{p+q}{2} \right)\cos \left( \frac{p-q}{2} \right)$

So $\sin \left( a+x \right)+\sin \left( \frac{\pi }{2}-x \right)=2\sin \left( \frac{a+x+\frac{\pi }{2}-x}{2} \right)\cos \left( \frac{a+x-\frac{\pi }{2}+x}{2} \right)$

                                    $=2\sin \left( \frac{a}{2}+\frac{\pi }{4} \right)\cos \left( \frac{a}{2}+x-\frac{\pi }{4} \right)$

So $\int_{0}^{\frac{\pi }{2}}{\frac{dx}{{{\left( \sin \left( a+x \right)+\cos x \right)}^{2}}}=\int_{0}^{\frac{\pi }{2}}{\frac{dx}{{{\left( 2\sin \left( \frac{a}{2}+\frac{\pi }{4} \right)\cos \left( \frac{a}{2}+x-\frac{\pi }{4} \right) \right)}^{2}}}}}$

$=\frac{1}{4{{\sin }^{2}}\left( \frac{a}{2}+\frac{\pi }{4} \right)}\int_{0}^{\frac{\pi }{2}}{{{\sec }^{2}}\left( \frac{a}{2}+x-\frac{\pi }{4} \right)dx}=\frac{1}{4{{\sin }^{2}}\left( \frac{a}{2}+\frac{\pi }{4} \right)}\int_{0}^{\frac{\pi }{2}}{d\left( \tan \left( \frac{a}{2}+x-\frac{\pi }{4} \right) \right)}$

$=\frac{1}{4{{\sin }^{2}}\left( \frac{a}{2}-\frac{\pi }{4} \right)}\tan \left( \frac{a}{2}+x-\frac{\pi }{4} \right)_{0}^{\frac{\pi }{2}}=\frac{1}{4{{\sin }^{2}}\left( \frac{a}{2}+\frac{\pi }{4} \right)}\left( \tan \left( \frac{a}{2}+\frac{\pi }{4} \right)-\tan \left( \frac{a}{2}-\frac{\pi }{4} \right) \right)$

But $\tan \left( \frac{a}{2}+\frac{\pi }{4} \right)-\tan \left( \frac{a}{2}-\frac{\pi }{4} \right)=\frac{\sin \left( \frac{a}{2}+\frac{\pi }{4} \right)}{\cos \left( \frac{a}{2}+\frac{\pi }{4} \right)}-\frac{\sin \left( \frac{a}{2}-\frac{\pi }{4} \right)}{\cos \left( \frac{a}{2}-\frac{\pi }{4} \right)}=\frac{\sin \left( \frac{\pi }{2} \right)}{\cos \left( \frac{a}{2}+\frac{\pi }{4} \right)\cos \left( \frac{a}{2}-\frac{\pi }{4} \right)}$

And ${{\sin }^{2}}\left( \frac{a}{2}+\frac{\pi }{4} \right)=\frac{1-\cos \left( 2\left( \frac{a}{2}+\frac{\pi }{4} \right) \right)}{2}=\frac{1-\cos \left( a+\frac{\pi }{2} \right)}{2}=\frac{1+\sin a}{2}$

Remark that $2\cos \left( \frac{p+q}{2} \right)\cos \left( \frac{p-q}{2} \right)=\cos \left( p \right)+\cos \left( q \right)$

So $\frac{p+q}{2}=\frac{a}{2}+\frac{\pi }{4}\,\,\,\,\And \,\,\,\,\frac{p-q}{2}=\frac{a}{2}-\frac{\pi }{4}$ $\Leftrightarrow 2p+2q=2a+\pi \,\,\,\And \,\,2p-2q=2a-\pi $

So $p=a\,\,\And \,\,\,q=\frac{\pi }{2}$ hence $\cos \left( \frac{a}{2}+\frac{\pi }{4} \right)\cos \left( \frac{a}{2}-\frac{\pi }{4} \right)=\frac{\cos \left( a \right)+\cos \left( \frac{\pi }{2} \right)}{2}=\frac{\cos a}{2}$

Thus $\int_{0}^{\frac{\pi }{2}}{\frac{dx}{{{\left( \sin \left( x+a \right)+\cos x \right)}^{2}}}=\frac{1}{4\frac{1+\sin a}{2}}\times \frac{1}{\frac{\cos a}{2}}=\frac{1}{\cos a\left( 1+\sin a \right)}}$